0
$\begingroup$

Hello I am doing practice problems and came across the following. The $i^{th}$ Rademacher function is defined on [0,1] as $R_i(x)= 2 a_i -1$ where a_i is the ith digit in the non terminating binary expansion of x. It is basically 1 if $a_i=1$ and -1 if $a_i=0$.

The problem is the following: Let $f_k(x)= \sum_{i=1}^{k} {\frac{1}{2i} R_i(x)}$ Compute $\int_{[0,1]} e^{t f_k(x)} dx$

I have no clue how to approach this problem, I thought about breaking it up into intervals but $R_k$ from 1 to k changes across any interval basically so that doesnt seem promising. Any clue how to approach this problem? Please thanks.

1 Answers 1

0

Let me look at this problem from a probabilistic point of view. If $X$ is uniformly distributed on $[0,\,1]$, and $a_i$ is the $i$th (random) digit in the binary expansion of $X$, then $R_i(X)$ are independent Rademacher r.v.'s with equally probable values $\pm 1$.

Then $f_k(X)=\sum_{i=1}^k \frac{1}{2i}R_i(X)$ is a sum of independent random variables.

We introduce random environment with the single goal: to express the required integral in more simple form.

The probability density function of $X$ is $1_{[0,\,1]}$: $$ f(x)=\begin{cases}1, & x\in[0,1]\cr 0 & x\not\in[0,1].\end{cases} $$ Remind that for r.v. $X$ with PDF $f(x)$ the expectation of any measurable function $g(X)$ is equal to $$Eg(X)=\int_{\mathbb R}g(x)f(x)\,dx.$$ Therefore $$Ee^{tfk(X)}=\int_{\mathbb R}e^{tfk(x)}f(x)\,dx = \int_0^1 e^{tfk(x)}\,dx. $$ This is the initial integral. So, we need to calculate $$\int_{[0,1]} e^{t f_k(x)} dx = Ee^{t f_k(X)}=Ee^{t\cdot \sum_{i=1}^k \frac{1}{2i}R_i(X)}=E\prod_{i=1}^k e^{\frac{t}{2i}R_i(X)}.$$

Independence of $R_i(X)$ over different $i$ implies $$ E\prod_{i=1}^k e^{\frac{t}{2i}R_i(X)} = \prod_{i=1}^k Ee^{\frac{t}{2i}R_i(X)}=\prod_{i=1}^k\left(\dfrac12 e^{\frac{t}{2i}}+\dfrac12 e^{-\frac{t}{2i}}\right)=\prod_{i=1}^k \cosh\left(\frac{t}{2i}\right). $$ Note that the last step we did by definition of expectation of discrete r.v.: for any function $h$ $$Eh(R_i)=h(1)\cdot P(R_i=1)+h(-1)\cdot P(R_i=-1)=\frac12 h(1)+\frac12 h(-1).$$ For $h(y)=e^{\frac{ty}{2i}}$ $$Ee^{\frac{tR_i}{2i}}=e^{\frac{t\cdot 1}{2i}}\cdot P\left(R_i=1\right)+e^{\frac{t\cdot (-1)}{2i}}\cdot P\left(R_i=-1\right)=\frac12 e^{\frac{t}{2i}} +\frac12 e^{-\frac{t}{2i}}.$$

  • 0
    Hi thanks, 2 quick questions, are we allowed to assume X is randomly distributed? And also although I follow the steps, I am a little confused how in the initial step we had to compute an integral and then a few steps later the expected value involved only a sum of two terms... Could you explain a bit more ? – 2017-02-22
  • 0
    We do not assume the initial $x$ is random. We take random $X$ and show that integral you a willing to calculate coincides with the expected value of some function of this r.v. $X$. For any measurable function $g$ and for continuously distributed $X$ with PDF $f(x)$, $Eg(X)=\int_{\mathbb R}g(x)f(x)\,dx$. We have applied this formula to $X$ uniformly distributed with PDF $f(x)=1$ on $[0,\,1]$ (and 0 otherwise), and $g(x)=e^{tf_k(x)}$. 2nd question: $R_i(X)=\pm 1$ with probability 1/2. So $Ee^{tR_i(X)/2i}=P(R_i(X)=1)e^{t\cdot 1/2i}+P(R_i(X)=-1)e^{t\cdot (-1)/2i}$. – 2017-02-22
  • 0
    Sorry for format mistakes in comments. I have added the answers to initial post. – 2017-02-22